1 00:00:14,210 --> 00:00:28,530 so let us look at a specific example so there is a composite slab and one end of this slab 2 00:00:28,530 --> 00:00:34,920 is maintained at no flux condition that is adiabatic conditions and the length is specified 3 00:00:34,920 --> 00:00:45,470 as fifty millimeters and twenty millimeters so thats the thickness of the slab and lets 4 00:00:45,470 --> 00:00:58,550 say that the q dot a and q dot b are the volumetric generation of heat in these two slabs and 5 00:00:58,550 --> 00:01:09,080 if k a and k b are the corresponding conductivities and if water is flowing outside ok water is 6 00:01:09,080 --> 00:01:16,620 flowing outside with at a temperature of thirty degree c and the heat transport coefficient 7 00:01:16,620 --> 00:01:24,680 is thousand watt per meter squared kelvin ok thats the heat transport coefficient and 8 00:01:24,680 --> 00:01:30,090 you can throw in some number so if the temperatures are specified as t 9 00:01:30,090 --> 00:01:40,890 naught t one and t two so if q dot a is one point five into ten power six watt per meter 10 00:01:40,890 --> 00:01:49,750 cube and k a thermal conductivity is given by seventy five watt per meter kelvin and 11 00:01:49,750 --> 00:01:55,930 if q dot b is zero which is which means that there is no heat generation and the second 12 00:01:55,930 --> 00:02:06,359 slab and if conductivity is given by one fifty watt per meter kelvin ok so we need to find 13 00:02:06,359 --> 00:02:19,260 the temperatures so the question is we need to find the temperatures of these three locations 14 00:02:19,260 --> 00:02:26,201 t naught is at the location where the adiabatic condition is maintained and t one is the interface 15 00:02:26,201 --> 00:02:30,880 between the two slabs and t two is the exterior ah exterior surface 16 00:02:30,880 --> 00:02:33,340 so how should we proceed 17 00:02:33,340 --> 00:02:41,820 yeah 18 00:02:41,820 --> 00:02:44,750 right balance equation for both can we do better than that 19 00:02:44,750 --> 00:02:50,340 hm 20 00:02:50,340 --> 00:02:54,280 we know the solution separately of course we can write the balances and we can find 21 00:02:54,280 --> 00:03:02,950 the solutions ok can we do it better than that so look at the system and intuit what 22 00:03:02,950 --> 00:03:09,010 whats the happening here what is happening here so you should not blindly go and write 23 00:03:09,010 --> 00:03:14,120 equations all the time you have to see what what the system is doing can we do better 24 00:03:14,120 --> 00:03:23,480 than that can we do in a simpler fashion can we solve the problem 25 00:03:23,480 --> 00:03:25,150 so what is happening at this interface 26 00:03:25,150 --> 00:03:30,970 yeah 27 00:03:30,970 --> 00:03:37,020 heat flux is same because whatever heat that is being transported from slab a is being 28 00:03:37,020 --> 00:03:51,040 transported to slab to at that interface ok what is the flux at that interface 29 00:03:51,040 --> 00:03:56,290 so note that this is adiabatic right so there is nothing that is leaving on this end so 30 00:03:56,290 --> 00:04:02,280 whatever heat that is being generated should actually be the flux at which the heat is 31 00:04:02,280 --> 00:04:08,010 being the total amount of heat that is generated because of the heat generation term must be 32 00:04:08,010 --> 00:04:14,680 transported only in one direction right so thats sort of obvious from the way the problem 33 00:04:14,680 --> 00:04:19,690 has been defined but you must intuit these kinds of things so so the total amount of 34 00:04:19,690 --> 00:04:32,620 heat that is generated so q dot a into the volume of this flat a 35 00:04:32,620 --> 00:04:38,610 which is lets say if this is l one length l one will put the numbers later 36 00:04:38,610 --> 00:04:47,720 so this is length l one l one into a so thats the total amount of heat that is generated 37 00:04:47,720 --> 00:04:52,150 and if we specify that its a steady state condition thats what we will be looking at 38 00:04:52,150 --> 00:04:59,300 that should be equal to the rate at which heat is being transported in slab b because 39 00:04:59,300 --> 00:05:04,560 there is no heat generation there so whatever comes in here has to go out if the steady 40 00:05:04,560 --> 00:05:15,580 state condition has to be maintained and that should be equal to h into t two minus t infinity 41 00:05:15,580 --> 00:05:23,800 right very simple so because there is constant heat transfer rate in slab b because q b dot 42 00:05:23,800 --> 00:05:29,590 is zero heat transfer rate is constant therefore whatever heat that is actually flowing at 43 00:05:29,590 --> 00:05:34,749 this interface whatever rate at which heat is being flowing from at this interface excuse 44 00:05:34,749 --> 00:05:40,610 me should be transported at this interface if you have to maintain steady state conditions 45 00:05:40,610 --> 00:05:47,639 therefore q dot into l one into a should be equal to h into a right so that is rate we 46 00:05:47,639 --> 00:05:56,969 are looking at rate not flux right so we know what q dot a is so this means q dot a into 47 00:05:56,969 --> 00:06:07,069 l one should be equal to h into t two minus t infinity so we know q dot a we know t infinity 48 00:06:07,069 --> 00:06:14,029 we know the heat transfer rate so we can find out what t two is right and so that will be 49 00:06:14,029 --> 00:06:24,889 q dot a into l one by h plus t infinity if i throw in the numbers it will be thirty plus 50 00:06:24,889 --> 00:06:33,199 two seventy three watch the units ok i am using s i units into ten power six multiplied 51 00:06:33,199 --> 00:06:44,430 by fifty into ten power minus three divided by thousand so thats approximately hundred 52 00:06:44,430 --> 00:06:50,639 and five plus two seventy three so that was one hundred and five degrees 53 00:06:50,639 --> 00:07:01,900 so thats the temperature t two ok is it clear to everyone how do we find t one 54 00:07:01,900 --> 00:07:10,310 how do we find t one we know that the heat transfer rate is constant and we can simply 55 00:07:10,310 --> 00:07:16,360 write resistance network right if we know how to write resistance network so the simply 56 00:07:16,360 --> 00:07:30,259 based on resistance network t one t two and that is given by l l two by k two into a right 57 00:07:30,259 --> 00:07:39,860 so thats the resistance network and this is q right so can we find t two from here 58 00:07:39,860 --> 00:07:50,650 no q b is zero there is no heat generation in the second slab if there is no heat generation 59 00:07:50,650 --> 00:07:58,430 in the seconds slab you would actually write the resistance network right so how do we 60 00:07:58,430 --> 00:08:10,509 solve this you dont know the area right so what do we do do we really need the area so 61 00:08:10,509 --> 00:08:16,719 how do we solve how do we draw the resistance network correct resistance network to find 62 00:08:16,719 --> 00:08:17,719 temperature t one 63 00:08:17,719 --> 00:08:23,759 area 64 00:08:23,759 --> 00:08:30,379 we know q equal to h into a right so so there are two ways of doing it you know what q is 65 00:08:30,379 --> 00:08:39,419 you could use this resistance network or you can draw overall resistance network also t 66 00:08:39,419 --> 00:08:45,970 infinity one call it t infinity and t one so you can use either of overall network or 67 00:08:45,970 --> 00:08:51,180 you know what q is and you can find out what what the temperature t one is ok so t one 68 00:08:51,180 --> 00:09:00,721 will be i will just put the number its very easy to calculate one fifteen degree c what 69 00:09:00,721 --> 00:09:17,680 about t zero how do we calculate t zero how do we calculate t zero yeah 70 00:09:17,680 --> 00:09:22,330 so we have to use the solution here we have to write the model if the model is not given 71 00:09:22,330 --> 00:09:27,810 to you obviously you have to write the model and find the solution because the resistance 72 00:09:27,810 --> 00:09:33,840 concept is not valid for that slab and therefore we have to write the model equation and find 73 00:09:33,840 --> 00:09:39,450 the solution and so i am not going to rewrite the equations here so we solve the equation 74 00:09:39,450 --> 00:09:47,240 just now so you can find out that ah q dot will be in sorry t naught will be t one plus 75 00:09:47,240 --> 00:09:56,520 q dot l square by two times k is equal to one hundred and forty degrees so remember 76 00:09:56,520 --> 00:10:07,490 that slab a is nothing but if you have if you have a slab which is twice the length 77 00:10:07,490 --> 00:10:13,570 of l one right so thats the problem that we just solved right so where the at the midpoint 78 00:10:13,570 --> 00:10:18,120 you have ah adiabatic condition right so thats the same problem so if you find what is the 79 00:10:18,120 --> 00:10:19,180 yes 80 00:10:19,180 --> 00:10:29,460 yeah sure so so remember that we solve this problem right at x equal to zero minus l n 81 00:10:29,460 --> 00:10:34,161 plus l and if you have a heat generation we solve this problem and we found the temperature 82 00:10:34,161 --> 00:10:39,740 profile correct and we know what is the temperature distribution what we need to know is what 83 00:10:39,740 --> 00:10:45,010 is the temperature at this location right thats the midpoint so remember that we also 84 00:10:45,010 --> 00:10:53,790 said solving this problem is equivalent to solving half slab starting from x equal to 85 00:10:53,790 --> 00:10:59,790 zero with the adiabatic condition why is it adiabatic because if you impose a symmetry 86 00:10:59,790 --> 00:11:05,410 condition on this where the temperatures on both sides are same then you have a maxima 87 00:11:05,410 --> 00:11:10,930 at the center so the profile looks like this so you have a maxima at the center which means 88 00:11:10,930 --> 00:11:16,110 that d t by d x is zero which is exactly the flux condition is exactly the flux boundary 89 00:11:16,110 --> 00:11:22,860 condition so this problem here is as good as solving a problem which is twice the size 90 00:11:22,860 --> 00:11:26,930 and if you get the temperature profile you can find out what is the temperature at the 91 00:11:26,930 --> 00:11:32,100 midpoint and thats what we solved a few moments ago 92 00:11:32,100 --> 00:11:36,380 so once you know the temperature this is simply given by this expression here and you will 93 00:11:36,380 --> 00:11:43,720 be able to find out so remember that the solution is given by i can give you what the solution 94 00:11:43,720 --> 00:11:47,400 is sir 95 00:11:47,400 --> 00:11:49,240 yes 96 00:11:49,240 --> 00:11:51,080 where 97 00:11:51,080 --> 00:11:52,390 yeah conduction is included in q b 98 00:11:52,390 --> 00:12:01,340 this is a balance at the interface the wall that is generated so not that the conduction 99 00:12:01,340 --> 00:12:08,080 is happening inside the slab right so at the boundary what comes out of the boundary is 100 00:12:08,080 --> 00:12:12,380 the total amount of heat that is generated inside the slab of course it comes through 101 00:12:12,380 --> 00:12:18,900 conduction how does heat come here because of conduction inside the slab but we are looking 102 00:12:18,900 --> 00:12:23,370 at what happens with that interface because there is no heat that is leaving on the other 103 00:12:23,370 --> 00:12:27,680 side everything has to leave through this side and so whatever comes here should be 104 00:12:27,680 --> 00:12:32,860 the total amount of heat that is transported the second slab because there is no heat generation 105 00:12:32,860 --> 00:12:46,700 in slab b clear so the solution of this equation would simply be t x is q dot l square by two 106 00:12:46,700 --> 00:12:54,381 k into one minus x square by l square plus t s so you put x equal to zero thats what 107 00:12:54,381 --> 00:13:01,590 you would get so t s is the temperature of the two surfaces 108 00:13:01,590 --> 00:13:06,420 which is same and that is nothing but p one and our problem here and so if you put x equal 109 00:13:06,420 --> 00:13:11,880 to zero here what you get is the solution here yes 110 00:13:11,880 --> 00:13:13,650 where 111 00:13:13,650 --> 00:13:20,190 you dont need to know you know what q is q is h into a into t two minus t infinity so 112 00:13:20,190 --> 00:13:25,320 the area will cancel out the area will cancel out when you actually write it its easy so 113 00:13:25,320 --> 00:13:33,460 we can write so t one maybe i will write it here for the benefit of everyone i will write 114 00:13:33,460 --> 00:13:44,510 it here so what is this so this is t one minus t two divided l two by k a k two a h into 115 00:13:44,510 --> 00:13:52,860 t two minus t infinity right so area will cancel out and you know what t two is already 116 00:13:52,860 --> 00:14:05,410 you know what t infinity so you should be able to find t one ok any other questions 117 00:14:05,410 --> 00:14:10,790 so i mean the take home message of this particular example the reason why i showed this is we 118 00:14:10,790 --> 00:14:15,950 should not blindly just go and write equations for a given problem you should first intuit 119 00:14:15,950 --> 00:14:22,180 what is happening in the problem can i make some estimate about certain quantities without 120 00:14:22,180 --> 00:14:26,850 solving the equations of course you can solve the equations and verify them but can i make 121 00:14:26,850 --> 00:14:32,240 some entity for example the intuitive judgment we made about heat transport a second slab 122 00:14:32,240 --> 00:14:37,020 such kind of judgments is very very important when you are handling these kinds of engineering 123 00:14:37,020 --> 00:14:40,830 problems so you must always pay attention and give 124 00:14:40,830 --> 00:14:47,740 utmost importance to your intuition ok so the next topic that we are going to see is 125 00:14:47,740 --> 00:14:53,010 extended surfaces so i will go to briefly give introduction to that in the next five 126 00:14:53,010 --> 00:15:06,050 minutes or so and then we will go deep into it in the next few lectures so so far we said 127 00:15:06,050 --> 00:15:15,220 by newtons law of cooling we said that q equal to h a into lets say the surface temperature 128 00:15:15,220 --> 00:15:21,290 minus t temperature of the fluid to which the heat is being transported so that k lets 129 00:15:21,290 --> 00:15:26,990 say it could be a slab like this where the fluid is actually flowing past the slab and 130 00:15:26,990 --> 00:15:32,910 this is the surface temperature and this is the temperature of the fluid now the question 131 00:15:32,910 --> 00:15:38,650 comes in most of the engineering problems is how can i increase the total amount of 132 00:15:38,650 --> 00:15:44,810 heat transport what are the ways by which how to increase 133 00:15:44,810 --> 00:15:58,420 its a very very important question how to increase q so the purpose of heat transport 134 00:15:58,420 --> 00:16:05,050 is basically to transport heat from one location to another so the better i can do the better 135 00:16:05,050 --> 00:16:11,660 placed i am right so if i can design a system which can transport heat better and obviously 136 00:16:11,660 --> 00:16:20,180 i am much better place so if you look at these three quantities here so if i call this as 137 00:16:20,180 --> 00:16:27,710 delta t ok some temperature difference so i can rewrite it as h into a into delta t 138 00:16:27,710 --> 00:16:35,450 so the net heat transfer rate depends upon three quantities right one is the heat transport 139 00:16:35,450 --> 00:16:42,070 coefficient the other one is the area of heat transport and the third one obviously is the 140 00:16:42,070 --> 00:16:46,530 temperature difference so these are the three quantities that needs to be tweaked so if 141 00:16:46,530 --> 00:16:51,220 you want to increase heat transfer rate you can increase either of them individually or 142 00:16:51,220 --> 00:16:55,190 you can increase multiple of them right many of them simultaneously if you are able 143 00:16:55,190 --> 00:16:59,490 to increase then you are going to increase the total amount of heat that is transported 144 00:16:59,490 --> 00:17:06,380 ok so when we discuss convection we are going to look at how to play with the heat transport 145 00:17:06,380 --> 00:17:11,390 coefficient what are the methods by which heat transport coefficient can be estimated 146 00:17:11,390 --> 00:17:16,010 and what are the factors on which the heat transfer coefficient depends upon and how 147 00:17:16,010 --> 00:17:22,179 to increase them is what we will see for most of the convection chapter ok so now what we 148 00:17:22,179 --> 00:17:28,529 are going to see is how to increase the area heat transport and thereby increase the net 149 00:17:28,529 --> 00:17:35,630 heat transfer rate so one way to do that is what is called the extended surfaces or in 150 00:17:35,630 --> 00:17:42,999 classical literature its also called as fins ok so there are several equipments where fins 151 00:17:42,999 --> 00:17:47,389 are being placed in order to increase the heat transport 152 00:17:47,389 --> 00:17:52,990 for example in the radiators in car where you want to dissipate the heat as soon as 153 00:17:52,990 --> 00:17:58,950 possible which means you need to have a very high heat transport so in order to increase 154 00:17:58,950 --> 00:18:05,490 the net heat transport you design a system which has very high surface area i in fact 155 00:18:05,490 --> 00:18:18,100 the way its done is so the fins sort of looks like this 156 00:18:18,100 --> 00:18:19,179 so they sort of look like this 157 00:18:19,179 --> 00:18:51,929 so so you have a base here ok so there is a base system which is maintained at a certain 158 00:18:51,929 --> 00:19:00,519 temperature so supposing if the temperature of the base surface is t s ok now you want 159 00:19:00,519 --> 00:19:06,340 to increase the heat transport from the base to the fluid which is circulating around ok 160 00:19:06,340 --> 00:19:12,280 so there is a fluid which is flowing here it could be any fluid in case of car radiators 161 00:19:12,280 --> 00:19:17,720 is basically air which is being circulated so you want to maximize the amount of heat 162 00:19:17,720 --> 00:19:22,530 that is transported from this surface to the fluid so the way to increase the heat transfer 163 00:19:22,530 --> 00:19:32,389 area is you create a thin protruding ah surfaces like this and so you have ah heat transfer 164 00:19:32,389 --> 00:19:37,840 occurring from the upper and the lower part of this protrusion that comes out of the surface 165 00:19:37,840 --> 00:19:43,379 and so you can have many such protrusions and each of these protrusion is what is called 166 00:19:43,379 --> 00:19:48,360 as a fin so you have now added some extra area for 167 00:19:48,360 --> 00:19:54,399 heat transport and thats how you enhance the net amount of heat that is being transported 168 00:19:54,399 --> 00:20:00,789 now if you look at it this is a slightly different problem than what we had dealt with so far 169 00:20:00,789 --> 00:20:08,600 so compare this kind of a structure with the structure that we had dealt with so far where 170 00:20:08,600 --> 00:20:15,080 the fluid which is being which is flowing is actually at the outer end of the system 171 00:20:15,080 --> 00:20:21,240 that you are dealing with and the flow of heat is actually in the direction perpendicular 172 00:20:21,240 --> 00:20:29,320 to that of the fluid flow right unlike here in this case the conduction is occurring in 173 00:20:29,320 --> 00:20:39,299 this direction while simultaneously there is heat loss from the from these surface right 174 00:20:39,299 --> 00:20:47,649 so here there is no simultaneous conduction of heat and heat loss from the surface while 175 00:20:47,649 --> 00:20:53,799 in the extended surface you have now introduced the system where there is simultaneous conduction 176 00:20:53,799 --> 00:20:59,690 of heat and also loss of heat to the fluid which is surrounding it so in this sense its 177 00:20:59,690 --> 00:21:03,950 a slightly a different problem and we are going to see how to account for these kinds 178 00:21:03,950 --> 00:21:09,350 of systems in fact that answers one of the questions we asked a short while ago what 179 00:21:09,350 --> 00:21:14,539 about heat lost from the system so here we are going to account for simultaneous loss 180 00:21:14,539 --> 00:21:21,210 of heat from the solid which is actually conducting heat in one direction so this is what we are 181 00:21:21,210 --> 00:21:26,470 going to see for the next couple of lectures and how to quantify heat transfer process 182 00:21:26,470 --> 00:21:31,480 and how to find out the net amount of heat that is transferred from these kinds of fins 183 00:21:31,480 --> 00:21:31,679 ok